1. If M is the sum of submodules N and L is it true that Ass(M) = Ass(N) ∪Ass(L)?Answer: No. Take A = Z, M = Z/2Z⊕Z, N = Z·(1, 1), and L = Z·(0, 1). ThenAss(M) = {2Z, 0}, but each of N and L is isomorphic to Z and has Ass = {0}.2. Let I = (x, y) ⊂ A = k[x, y, z]/(xy − z2). Find Ass(A/I2). (Optional: find aprimary decomposition of I2).Answer: We have (x, y, z) ⊆ rad(I2), since x2= x2∈ I2, y2= y2∈ I2, z2=xy ∈ I2. Since (x, y, z) is maximal and I26= (1), this means we have equality. Since(x, y, z) is maximal, we get that I2is (x, y, z)-primary, so its Ass is just {(x, y, z)},and it is its own primary decomposition.3. Let A be Noetherian and M, N finite modules over A. Show that Ass(Hom(M, N)) =Supp(M) ∩ Ass(N).Proof: Man, you guys really need to learn how to work with the commutativealgebra instead of against it. It’s a Tao thing, wei wu wei. Here are the lemmaswhich make the problem trivial (the first one could probably be used to shortenthe exposition of all this Ass stuff):Lemma 1: Let A be a ring, M an A-module. ThenAss(M) = {P ∈ Spec(A) | Hom(A/P, M )P6= 0}.Proof: We have Hom(A/P, M )P6= 0 if and only if there is a φ ∈ Hom(A/P, M)with Ann(φ) ⊆ P . But giving a φ ∈ Hom(A/P, M ) is exactly the same as giving anelement (namely φ(1)) of M annihilated by P ; and the annihilator of φ is then thesame as the annihilator of that element; so the condition on this element is that ithave annihilator both contained in and containing P . Lemma 2: Let A be a ring, B a flat A-algebra (e.g. B = APfor some primeP ), M a finitely presented A-module (this is equivalent to finitely generated if A isNoetherian), and N an arbitrary A-module. Then there’s a natural isomorphismof B-modulesHomA(M, N)B∼−→ HomB(MB, NB).Proof: First we give the map: send φ ⊗ b to m ⊗ b07→ φ(m) ⊗ (bb0). This isclearly natural, well-defined, B-linear, etc. etc.Now we check that it’s an isomorphism in the special case M = Anfor somen ≥ 0. Well, in that case, HomA(M, N)∼=Nn, and HomB(MB, NB)∼=(NB)n, andit’s easy to see that via these identifications, the above map becomes the obviousmap (Nn)B→ (NB)n, which we know to be an isomorphism.For the general case, since M is finitely-presented, we have an exact sequenceAm→ An→ M → 0.To this we can apply the functors HomA(−, N)Band HomB((−)B, NB), both ofwhich are contravariant left-exact by general nonsense and the flatness hypothesis;12then we get a commutative diagram where the rows are exact, the vertical arrowsall instances of the above-defined map:0//HomA(M, N)B//HomA(An, N)B//HomA(Am, N)B0//HomB(MB, NB)//HomB((An)B, NB)//HomB((Am)B, NB).Then by the first thing we checked, the last two vertical maps are isomorphisms;by diagram chasing, then, so is the first one, as desired. OK, armed with our lemmas, let’s just compute. To find Ass(Hom(M, N )), bythe first lemma, we should considerHom(A/P, Hom(M, N))P∼=Hom(M, Hom(A/P, N))P∼=HomAP(MP, Hom(A/P, N)P)∼=Homk(MP⊗APk, Hom(A/P, N)P).Here k = AP/P AP, the first isomorphism is because both sides are just bilin-ear maps A/P × M → N localized at P , the second isomorphism is the sec-ond lemma, and the third isomorphism is because Hom(A/P, N )Pis killed by P ,and is thus a k-module. But then look at what we have: we’re working overa field, so it’s trivial that this last thing is 0 if and only if MP⊗APk = 0 orHom(A/P, N)P= 0. By Nakayama (again we use M finite), this is the same asMP= 0 or Hom(A/P, N)P= 0. By the first lemma, we’re done. We see that thestatement holds for general A, finitely presented M, and arbitrary N . 3.16. Let B be a flat A-algebra. TFAE:(1) Ice= I for all ideals I ⊆ A.(2) Spec(B) → Spec(A) is surjective.(3) For every maximal ideal m of A we have me6= (1).(4) If M 6= 0 is an A-module, then MB6= 0.(5) For every A-module M , the natural M → MBis injective.Proof: (1) ⇒ (2) is one of the problems on the last set.(2) ⇒ (3): Say m = Pcfor P ⊆ B prime. Then me= Pce⊆ P ( (1).(3) ⇒ (4): Let m ∈ M with m 6= 0. The map A → M given by a 7→ a · m haskernel I = Ann(m) 6= (1), so we get an exact0 → A/I → M.3Tensoring gives, by flatness and one of our first exercises,0 → B/IB → MB.But if I ⊆ m, then IB ⊆ me( (1), so B/IB 6= 0, so MB6= 0.(4) ⇒ (5): Let N be the kernel, so0 → N → M → MB.Flatness gives0 → NB→ MB→ MB⊗AB.However, this last map has a left inverse, namely (m ⊗ b) ⊗ b07→ m ⊗ (bb0), whichmeans it’s injective, so NB= 0, so N = 0, as desired.(5) ⇒ (1): Take M = A/I; we get that A/I → B/IB is injective, which exactlysays I = Ice.4.2. Let I be a radical ideal in a Noetherian ring A. Then I has no embeddedprimes.Proof: Recall Spec(A/I) is noetherian, so it’s a union of finitely many irre-ducible components, which correspond to minimal primes of A/I. Radical idealscorresponding to closed subsets of Spec, we get that I is the intersection of the(finitely many) primes minimal over I. This is a primary decomposition, and byminimality none are